The condition that allows gluing of symplectic manifolds, is the existence of a fixed point free $S^1$ action on the boundary, such that the orbits of the $S^1$ action are tangent to the kernel of the restriction of the symplectic form to the boundary, $\ker(\omega|_W)$.

If two symplectic manifolds have orientation-reversing diffeomorphic boundaries, each with an $S^1$ action as above, and the symplectic reductions of the boundaries are symplectomorphic, then the manifolds can be glued together to get a closed symplectic manifold. The symplectic reduction here is the quotient of the boundary by the $S^1$ action, with the induced symplectic form (which remains non-degenerate since we are modding out by the kernel directions).

One significant difference between convex/concave gluing and this gluing is that one is inherently directional — a symplectic manifold with convex boundary cannot be viewed as a symplectic manifold with concave boundary by simply reversing some orientations, or negating the symplectic structure (and vice versa). So to glue, we can’t start with two symplectic manifolds which both have convex ends. Symplectic manifolds whose boundaries have $S^1$ action satisfying the correct conditions, are similarly considered either positive or negative, and to glue them together we need one postive side and one negative side. However it is possible to take a positive side and turn it into a negative side simply by reversing the orientation of the $S^1$ action, so we can glue two positive sides together simply by reversing the $S^1$ action on one of them to turn it into a negative side.

Theorem [McDuff (free action case), McCarthy-Wolfson (fixed point free generalization), converse is Duistermaat-Heckman]: If I is an interval of regular values containing c, the symplectic form on $H^{-1}(I)$ is uniquely determined by the manifold $N=H^{-1}(c)$, the $S^1$ action on $N$, and a family of symplectic forms $\{\tau_t\}$ on the (orbifolds) $H^{-1}(t)/S^1$, where $\tau_{t_2}-\tau_{t_1}=(t_1-t_2)c$ ($c$ is the Chern class of a principal $S^1$ bundle over the orbifold which is finitely branched covered by $N$ — in the case the circle action on $N$ is free, this principal bundle is just $N$).

The idea is that the symplectic forms $\tau_t$ are the symplectic reduction of \omega on $N/S^1$, and the symplectic form in the other directions are determined by the $S^1$ invariance. It turns out that this is the model for any hypersurface of a symplectic manifold with a compatible $S^1$ action. If N is any hypersurface with a fixed point free $S^1$ action such that $\ker(\omega|N)$ is tangent to the orbits, then there exists a neighborhood of N, symplectomorphic to a standard model: $(N\times (-\varepsilon,\varepsilon),\omega_0)$ coming from a Hamiltonian circle action. Call an embedding of $N$ into $(M,\omega)$, $\omega$-compatible if the image of the $S^1$ orbits is tangent to $\ker(\omega|_N)$.

McCarthy and Wolfson prove and use this theorem to show that one can cut two closed symplectic manifolds along such an $N$, and switch the top pieces and reglue to get two new symplectic manifolds. The “top” or + piece is determined as follows. The circle action is oriented and is tangent to the vector field $v_H$. Since the symplectic form on $(M,\omega)$ is non-degenerate, and $v_H\in \ker(\omega|_N)$, there must be a vector field $X$ transverse to $N$, which pairs positively with $v_H, \; \omega(v_H,X)>0$. The piece of the complement of N for which this vector field $X$ points inward is the top, $M^+$ side, and the piece for which this vector field points outward is the bottom, $M^-$ side. McCarthy and Wolfson state their gluing theorem as follows:

Theorem: Suppose $(M_i^{2n},\omega_i)$ for $i=1,2$ are symplectic manifolds, $N^{2n-1}$ is a manifold with a fixed point free $S^1$ action, and there are $\omega_i$-compatible embeddings $j_i: N\to (M_i\omega_i)$ such that the image of $N$ is a separating hypersurface. Also suppose that the symplectic reductions $(j_i(N)/S^1,\pi_*(\omega|{j_i(N)}))$ are symplectomorphic. Then there is a symplectic structure \omega on $M=M_1^-\cup_N M_2^+$, which agrees with the original symplectic structures, on a complement of a neighborhood of $N$.

At initial glance, one would think that the $M^-$ and $M^+$ sides correspond to the concave and convex fillings. However, it is not hard to show that there is nothing dictating that the $M^-$ side cannot act as an $M^+$ side. By simply negating the Hamiltonian, $H\mapsto -H$, we reverse the Hamiltonian vector field $v_H\mapsto -v_H$, and thus the $S^1$ action. This does not change whether the hypotheses of the theorem are satisfied: the $S^1$ action with reversed orientation is still tangent to $\ker(\omega|{j_i(N)}$), and the symplectic reductions are unchanged. However if before $\omega(v_H,X)>0$ and $X$ pointed into $M^+$ and out of $M^-$, then now $\omega(-v_H,-X)>0$, and $-X$ points into $M^-$ and out of $M^+$, so $M^+$ and $M^-$ have exchanged roles. In other words, the property of being a symplectic manifold whose boundary has a fixed point free $S^1$ is not an inherently sided/directional condition like a boundary of contact type is. The + and – sides look the same after a $180^{\circ}$ rotation of the vectors giving the circle and normal directions.

With this in mind, I do not see why this technique can not be applied to glue a $4$-ball to itself across $S^3$?

It looks like all the conditions are satisfied ... but the gluing seems impossible because of the lack of symplectic structure on the glued manifold... Any help will be highly appreciated...

  • $\begingroup$ Where can I find more about this gluing construction of manifolds with an $\mathbb{S}^1$-action? $\endgroup$ Feb 25 '19 at 23:27

$S^3$ as a boundary cannot be both convex and concave. This is proved in the famous paper of Eliashberg-Gromov.

  • 2
    $\begingroup$ Dear YHBKJ, the gluing specified above is not a concave/convex gluing. Please read the question carefully. $\endgroup$
    – RAG
    Feb 2 '17 at 16:48
  • $\begingroup$ @RAG Have you read the paper of McCarthy-Wolfson carefully? This is a special case of convex-concave gluing. $\endgroup$
    – YHBKJ
    Feb 2 '17 at 17:12

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